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• Validating Login / Changing User settings / Php Mysql

  - by Marcelo

  Hi everyone, my questions are about login, and changing already saved data. (Q1) 'Till now I've only saved input in the tables of the database (registration steps), now I need to check if the input (login steps), are the same of my table in database, in fact I have 3 types of users, then I'll have to check 3 kind of tables. Then if the input data matches with one of those 3 tables I will redirect the user to his specific area. I'm thinking about saved the submitted data $login=$_REQUEST['login']; and $password=$_REQUEST['password']; and compare with the login column in the database. Then if the login matches, I'll compare the password submitted with the one in the row, not in the column. But I don't know how to do this search and comparison,neither what to use. Then if both matches I'll redirect the user. Else I'll send an login error message. (this I know how to do) (Q2) What if need to change an already saved user ? For example to change an email address. My changing user's data web page is exactly the same like the registration user web page. Can I load the already saved options and values of registration (table user for example). Then the user will change whatever he thinks it's necessary, and then when he submits the new information, they would not create a new row in my table, but just be overwritten the old information? How can I do this? Sorry for any mistake in English, and Thanks for the attention.

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• ASP.NET Response Filter to Reformat the rendered output of ASPX pages?

  - by PropellerHead

  I've created a simple HttpModule and response stream to reformat the rendered output of web pages (see code snippets below). In the HttpModule I set the Response.Filter to my PageStream: m_Application.Context.Response.Filter = new PageStream(m_Application.Context); In the PageStream I overwrite the Write method in order to do my reformatting of the rendered output: public override void Write(byte[] buffer, int offset, int count) { string html = System.Text.Encoding.UTF8.GetString(buffer); //Do some string resplace operations here... byte[] input = System.Text.Encoding.UTF8.GetBytes(html); m_DefaultStream.Write(input, 0, input.Length); } And this work fine when using it on simple HTML pages (.html), but when I use this method on ASPX pages (.aspx), the Write method is called several times, splitting up the reformatting into different steps, and potentially destroying the string replacement operations. How do I solve this? Is there a way to let the ASPX page NOT call Write several times, e.g. by changing its buffer size, or have I chosen the wrong approach entirely, by using this Response.Filter method to manipulate the rendered output?

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• Codeigniter: Simple ajax submit not working

  - by Kevin Brown

  jQuery: $('#welcome-message').submit(function() { // inside event callbacks 'this' is the DOM element so we first // wrap it in a jQuery object and then invoke ajaxSubmit $(this).ajaxSubmit({}); $(this).fadeTo(400, 0, function () { // Links with the class "close" will close parent $(this).slideUp(400); }); return false; }); Controller: function welcome_message() { if(isset($_POST['welcome_message'])) { $this-_my_private_function(); } } private function _my_private_function() { $id = $this->session->userdata('id'); $profile['welcome_message'] = '0'; $this->db->update('be_user_profiles',$profile, array('user_id' => $id)); redirect('home', 'location'); } html: <?php print form_open('home/welcome_message',array('class'=>'horizontal','id'=>'welcome-message'))?> <p> Before you can complete the assessment, you need to complete your profile. Once that's done you'll be ready! After you have completed the assessment, you will be able to view the results from your profile. </p> <input type="checkbox" value="0" name="welcome_message" checked="false"> Don't show me this again </input> <p> <input class="button submit" type="submit" class="close-box" value="Close" /> </p> <?php print form_close()?> This should be working. My hypothesis is that the data isn't being passed to the function...but I really have no idea! It works when I visit the function in the url.

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• Vertically Aligning Divs

  - by Crays

  Hi, i'm trying to make a small username and password input box. I would like to ask how do you vertically align a div? What i have is a <div id="Login" class="BlackStrip floatright"> <div id="Username" class="floatleft">Username<br>Password</div> <div id="Form" class="floatleft"> <form action="" method="post"> <input type="text" border="0"><br> <input type="password" border="0"> </form> </div> </div> and how can i make the div with id Username and Form to vertically align itself to the center? I tried putting vertical-align: middle; in css for the div with id Login. But doesn't seem to work. Any help would be appreciated. Thanks.

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• Location of Embedly (JQuery-Preview) Results

  - by user749798

  Embedly/Jquery-Preview has been fantastic. However, I'm trying to change the location of the preview result, and having trouble. Right now, the result appears right below the input field...but I'd rather have it in a separate part of the page. Is there a way to do this? I've tried changing the location of the selector and loading divs, but that hasn't helped. It seems to ignore those divs and put it right below the submit button. Below is my code: <form accept-charset="UTF-8" action="private" class="new_comment" data-remote="true" id="new_comment" method="post"> <input class="photo_comm" id="comment_comment" name="comment[comment]" placeholder="add a comment or link..." size="30" type="text" /><span type="text" id="counter">1000</span> <input class="btn btn-primary btn-mini" data-disable-with="Submitting..." name="commit" type="submit" value="Post" /> </form> <!-- Placeholder that tells Preview where to put the loading icon--> <div class="loading"> <img src='http://embedly.github.com/jquery-preview/images/loading-rectangle.gif'> </div> <!-- Placeholder that tells Preview where to put the selector--> <div class="selector"></div> $('#comment_comment').preview({ key:'60f1dcdf3258476794784148a6eb65e7', // Sign up for a key: http://embed.ly/pricing selector : {type:'rich'}, preview : { submit : function(e, data){ e.preventDefault(); $.ajax({ dataType: 'script', url: this.form.attr('action'), type: 'POST', data: data }); }, }, autoplay : 0, maxwidth : 400, display : {display : 'rich'} });

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• ArrayList<Int> Collections.Sort and LineNumberReader Help How to

  - by user1819551

  I have a issue i can't get it to work now let going to the point a explain in the code thanks. This is my class: what I want to do is insert the Integers sort the list and buffer writer in a column with out coma. Now I getting this: [1110018, 1110032, 1110056, 1110059, 1110063, 1110085, 1110096, 1110123, 1110125, 1110185, 1110456, 1110459] I want like this: 111xxxxx 111xxxx xxxx....... I can't do it in single array, have to be in ArrayList. This is my collecting: list.addNumbers(numbers); list.display(); This is my writer: Is buffered coma.write("\n"+list.display()); coma.flush();<br/> Here is my class: public class IdCount {<br/> private ArrayList<Integer> properNumber = new ArrayList<>(); public void addNumbers(Integer numbers) { properNumber.add(numbers); Collections.sort(properNumber); } public String display() { //(I try .toString() Not work) return properNumber.toString(); } My second issue is LineNumberReader: This is my collecting and my writing: try { Reader input = new BufferedReader( new FileReader(inputFile)); try (Scanner in = new Scanner(input)) { while (in.hasNext()) { //(More Code) asp = new LineNumberReader(input); int rom = 0; while (asp.readLine()!=null){ rom++; } System.out.println(rom); coma.write(rom); This one will not write anything an my System Print give me only 12 0 in column.

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• PHP/MySQL allowing current user to edit there account information

  - by user1837896

  i have created 2 pages update.php edit.php we start on edit.php so here is edit.php's script <?php session_start(); $id = $_SESSION["id"]; $username = $_POST["username"]; $fname = $_POST["fname"]; $password = $_POST["password"]; $email = $_POST["email"]; mysql_connect('mysql13.000webhost.com', 'a2670376_Users', 'Password') or die(mysql_error()); echo "MySQL Connection Established! <br>"; mysql_select_db("a2670376_Pass") or die(mysql_error()); echo "Database Found! <br>"; $query = "UPDATE members SET username = '$username', fname = '$fname', password = '$password' WHERE id = '$id'"; $res = mysql_query($query); if ($res) echo "<p>Record Updated<p>"; else echo "Problem updating record. MySQL Error: " . mysql_error(); ?> <form action="update.php" method="post"> <input type="hidden" name="id" value="<?=$id;?>"> ScreenName:<br> <input type='text' name='username' id='username' maxlength='25' style='width:247px' name="username" value="<?=$username;?>"/><br> FullName:<br> <input type='text' name='fname' id='fname' maxlength='20' style='width:248px' name="ud_img" value="<?=$fname;?>"/><br> Email:<br> <input type='text' name='email' id='email' maxlength='50' style='width:250px' name="ud_img" value="<?=$email;?>"/><br> Password:<br> <input type='text' name='password' id='password' maxlength='25' style='width:251px' value="<?=$password;?>"/><br> <input type="Submit"> </form> now here is the update.php page where i am having the MAJOR problem <?php session_start(); mysql_connect('mysql13.000webhost.com', 'a2670376_Users', 'Password') or die(mysql_error()); mysql_select_db("a2670376_Pass") or die(mysql_error()); $id = (int)$_SESSION["id"]; $username = mysql_real_escape_string($_POST["username"]); $fname = mysql_real_escape_string($_POST["fname"]); $email = mysql_real_escape_string($_POST["email"]); $password = mysql_real_escape_string($_POST["password"]); $query="UPDATE members SET username = '$username', fname = '$fname', email = '$email', password = '$password' WHERE id='$id'"; mysql_query($query)or die(mysql_error()); if(mysql_affected_rows()>=1){ echo "<p>($id) Record Updated<p>"; }else{ echo "<p>($id) Not Updated<p>"; } ?> now on edit.php i fill out the form to edit the account "test" while i am logged into it now once the form if filled out i click on |Submit!| button and it takes me to update.php and it returns this (0) Not Updated (0) <= id of user logged in Not Updated <= MySql Error from mysql_query($query)or die(mysql_error()); if(mysql_affected_rows()>=1){ i want it to update the user logged in and if i am not mistaken in this script it says $id = (int)$_SESSION["id"]; witch updates the user with the id of the person who is logged in but it isnt updating its saying that no tables were effected if it helps heres my MySql Database picture just click here http://i50.tinypic.com/21juqfq.png if this could possibly be any help to find the solution i have 2 more files delete.php and delete_ac.php they have can remove users from my sql database and they show the user id and it works there are no bugs in this script at all PLEASE DO NOT MAKE SUGGESTIONS FOR THE SCRIPTS BELOW delete.php first <?php $host="mysql13.000webhost.com"; // Host name $username="a2670376_Users"; // Mysql username $password="PASSWORD"; // Mysql password $db_name="a2670376_Pass"; // Database name $tbl_name="members"; // Table name // Connect to server and select database. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); // select record from mysql $sql="SELECT * FROM $tbl_name"; $result=mysql_query($sql); ?> <table border="0" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC"> <tr> <td colspan="8" style="bgcolor: #FFFFFF"><strong><img src="http://i47.tinypic.com/u6ihk.png" height="30" widht="30">Delete data in mysql</strong> </td> </tr> <tr> <td align="center" bgcolor="#FFFFFF"><strong>Id</strong></td> <td align="center" bgcolor="#FFFFFF"><strong>UserName</strong></td> <td align="center" bgcolor="#FFFFFF"><strong>FullName</strong></td> <td align="center" bgcolor="#FFFFFF"><strong>Password</strong></td> <td align="center" bgcolor="#FFFFFF"><strong>Email</strong></td> <td align="center" bgcolor="#FFFFFF"><strong>Date</strong></td> <td align="center" bgcolor="#FFFFFF"><strong>Ip</strong></td> <td align="center" bgcolor="#FFFFFF">&nbsp;</td> </tr> <?php while($rows=mysql_fetch_array($result)){ ?> <tr> <td bgcolor="#FFFFFF"><? echo $rows['id']; ?></td> <td bgcolor="#FFFFFF"><? echo $rows['username']; ?></td> <td bgcolor="#FFFFFF"><? echo $rows['fname']; ?></td> <td bgcolor="#FFFFFF"><? echo $rows['password']; ?></td> <td bgcolor="#FFFFFF"><? echo $rows['email']; ?></td> <td bgcolor="#FFFFFF"><? echo $rows['date']; ?></td> <td bgcolor="#FFFFFF"><? echo $rows['ip']; ?></td> <td bgcolor="#FFFFFF"><a href="delete_ac.php?id=<? echo $rows['id']; ?>">delete</a></td> </tr> <?php // close while loop } ?> </table> <?php // close connection; sql_close(); ?> and now delete_ac.php <table width="500" border="0" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC"> <tr> <td colspan="8" bgcolor="#FFFFFF"><strong><img src="http://t2.gstatic.com/images? q=tbn:ANd9GcS_kwpNSSt3UuBHxq5zhkJQAlPnaXyePaw07R652f4StmvIQAAf6g" height="30" widht="30">Removal Of Account</strong> </td> </tr> <tr> <td align="center" bgcolor="#FFFFFF"> <?php $host="mysql13.000webhost.com"; // Host name $username="a2670376_Users"; // Mysql username $password="javascript00"; // Mysql password $db_name="a2670376_Pass"; // Database name $tbl_name="members"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); // get value of id that sent from address bar $id=$_GET['id']; // Delete data in mysql from row that has this id $sql="DELETE FROM $tbl_name WHERE id='$id'"; $result=mysql_query($sql); // if successfully deleted if($result){ echo "Deleted Successfully"; echo "<BR>"; echo "<a href='delete.php'>Back to main page</a>"; } else { echo "ERROR"; } ?> <?php // close connection mysql_close(); ?> </td> </tr> </table>

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• Reading only one row from table in model binder

  - by user281180

  I am filling a table dynamically. I can see the table filled with 3 rows, but in my model binder I can read only one value. How can I solve this problem? My code is as follows: function AddTableRow(jQtable, value, text){ var count = 0; jQtable.each(function() { var $table = $(this); var tds = '<tr>'; tds += '<td>' + '<input type="text" value = ' + text + ' disabled ="disabled" style="width:auto"/>' + '<input type="hidden" name="projectList[' + count + '].ID" value = ' + value + ' /></td>' + '<td><input type="button" value="Remove"/></td>'; tds += '</tr>'; if ($('tbody', this).length > 0) { $('tbody', this).append(tds); } else { $(this).append(tds); } count++;}); } function ReadSelectedProject() { $("#Selected option").each(function() { AddTableRow($('#projectTable'), $(this).val(), $(this).text()); }); }

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• sigleton EXC_BAD_ACCESS

  - by lclaud

  Hi, so I have a class that I decleare as a singleton and in that class I have a NSMutableArray that contains some NSDictionaries with some key/value pairs in them. The trouble is it doesn't work and I dont't know why... I mean it crashes with EXC_BAD_ACCESS but i don't know where. I followed the code and it did create a new array on first add, made it to the end of the funtion ..and crashed ... @interface dataBase : NSObject { NSMutableArray *inregistrari; } @property (nonatomic,retain) NSMutableArray *inregistrari; -(void)adaugaInregistrareCuData:(NSDate *)data siValoare:(NSNumber *)suma caVenit:(BOOL)venit cuDetaliu:(NSString *)detaliu; -(NSDictionary *)raportIntreData:(NSDate *)dataInitiala siData:(NSDate *)dataFinala; -(NSArray *)luniDisponibileIntreData:(NSDate *)dataInitiala siData:(NSDate *)dataFinala; -(NSArray *)aniDisponibiliIntreData:(NSDate *)dataInitiala siData:(NSDate *)dataFinala; -(NSArray *)vectorDateIntreData:(NSDate *)dataI siData:(NSDate *)dataF; -(void)salveazaInFisier; -(void)incarcaDinFisier; + (dataBase *)shareddataBase; @end And here is the .m file #import "dataBase.h" #import "SynthesizeSingleton.h" @implementation dataBase @synthesize inregistrari; SYNTHESIZE_SINGLETON_FOR_CLASS(dataBase); -(void)adaugaInregistrareCuData:(NSDate *)data siValoare:(NSNumber *)suma caVenit:(BOOL)venit cuDetaliu:(NSString *)detaliu{ NSNumber *v=[NSNumber numberWithBool:venit]; NSArray *input=[NSArray arrayWithObjects:data,suma,v,detaliu,nil]; NSArray *keys=[NSArray arrayWithObjects:@"data",@"suma",@"venit",@"detaliu",nil]; NSDictionary *inreg=[NSDictionary dictionaryWithObjects:input forKeys:keys]; if(inregistrari == nil) { inregistrari=[[NSMutableArray alloc ] initWithObjects:inreg,nil]; }else { [inregistrari addObject:inreg]; } [inreg release]; [input release]; [keys release]; } It made it to the end of that adaugaInregistrareCuData ... ok . said the array had one object ... and then crashed

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• how to send on users profile page on selecting username( using json autosuggest script)

  - by I Like PHP

  i m using auto suggest using Ajax Json . now when a user select a user name , i want to send user on the link of that user name my json data is coming in this way { query:'hel', suggestions:["hello world","hell boy ","bac to hell"], data:["2","26","34"] } now what i want that user goes to http://userProfile.php?uid=26 on select username(suppose user select "hell boy") how to do this?? UPDATE: i describe what i m doing step by step i m using a searchbox using jquery ajax, when user write some text on input box , we show (suggest) value regarading their search STEP 1. when user write atleast(2 letter) <input type="text" name="q" id="query" />then a function(below) in invoked in which i send the value written on text box(eg. hell). <script type="text/javascript" language="javascript"> var options, a; jQuery(function(){ options = { serviceUrl:'rpc.php' }; var a = $('#query').autocomplete({ serviceUrl:'rpc.php', minChars:3, delimiter: /(,|;)\s*/, // regex or character maxHeight:400, width:300, zIndex: 9999, deferRequestBy: 0, //miliseconds }); }); </script> STEP 3: on rpc.php, i collect the data and show using JSON my final data come in below format { query:'hell', suggestions:["hello world","hell boy ","bac to hell"], data:["2","26","34"] } where suggestion list having username and data is userid( from user_tables). above data comes in a div (on frontend) where user name displayed in a list STEP 4: now if i select any username using uparrow, downarrow then that name is filled in input box, STEP 5: now what i want that when user select usename then page automatically goes to that user's profile section ( userprofile.php?uid=2)

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• How to insert an element between the two elements dynamically?

  - by Harish

  I am using a table, in which there are buttons, on button click i want the new TR element to be inserted between the two TR or at the end of the TR... my code goes here <table> <tbody> <tr> <td> <input type="submit" value="Add" onclick="addFunction()" /> </td> </tr> <tr> <td> <input type="submit" value="Add" onclick="addFunction()" /> </td> </tr> <tr> <td> <input type="submit" value="Add" onclick="addFunction()" /> </td> </tr> </tbody> </table> i want to insert new TR element next to the element which has triggered the event... NOTE: i am not using any javascript library, just plain javascript

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• adding an uncertain number of fields using javascript

  - by user306472

  I'm new to javascript and a novice programmer, so this might be a really easy question to answer. I would like to loop over the values of x number of fields and add them up to display the sum in a final field. I can perform this function if I explicitly call each field, but I want to abstract this so I can deal with a flexible number of fields. Here's example code I've come up with (that's not working for me). Where am I going wrong? <html> <head> <script type="text/javascript"> function startAdd(){ interval = setInterval("addfields()",1); } function addfields(){ a = document.addition.total.value; b = getElementByTagName("sum").value; for (i=0; i<=b.length; i++){ a+=i; } return a; } function stopAdd(){ clearInterval(interval); } </script> </head> <body> <form name="addition"> <input type="text" name="sum" onFocus="startAdd();" onBlur="stopAdd;"> + <input type="text" name="sum" onFocus="startAdd();" onBlur="stopAdd;"> = <input type="text" name ="total"> </form> </body> </html>

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• Why do the outputs differ when I run this code using Netbeans 6.8 and Eclipse?

  - by Vimal Basdeo

  When I am running the following codes using Eclipse and Netbeans 6.8. I want to see the available COM ports on my computer. When running in Eclipse it is returning me all available COm ports but when running t in Netbeans, it does not seem to find any ports .. public static void test(){ Enumeration lists=CommPortIdentifier.getPortIdentifiers(); System.out.println(lists.hasMoreElements()); while (lists.hasMoreElements()){ CommPortIdentifier cn=(CommPortIdentifier)lists.nextElement(); if ((CommPortIdentifier.PORT_SERIAL==cn.getPortType())){ System.out.println("Name is serail portzzzz "+cn.getName()+" Owned status "+cn.isCurrentlyOwned()); try{ SerialPort port1=(SerialPort)cn.open("ComControl",800000); port1.setSerialPortParams(9600, SerialPort.DATABITS_8, SerialPort.STOPBITS_1, SerialPort.PARITY_NONE); System.out.println("Before get stream"); OutputStream out=port1.getOutputStream(); InputStream input=port1.getInputStream(); System.out.println("Before write"); out.write("AT".getBytes()); System.out.println("After write"); int sample=0; //while((( sample=input.read())!=-1)){ System.out.println("Before read"); //System.out.println(input.read() + "TEsting "); //} System.out.println("After read"); System.out.println("Receive timeout is "+port1.getReceiveTimeout()); }catch(Exception e){ System.err.println(e.getMessage()); } } else{ System.out.println("Name is parallel portzzzz "+cn.getName()+" Owned status "+cn.isCurrentlyOwned()+cn.getPortType()+" "); } } } Output with Netbeans false Output using Eclipse true Name is serail portzzzz COM1 Owned status false Before get stream Before write After write Before read After read Receive timeout is -1 Name is serail portzzzz COM2 Owned status false Before get stream Before write After write Before read After read Receive timeout is -1 Name is parallel portzzzz LPT1 Owned status false2 Name is parallel portzzzz LPT2 Owned status false2

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• How to upload files to server using JSP/Servlet?

  - by Thang Pham

  How can I upload files to server using JSP/Servlet? I tried this: <form action="upload" method="post"> <input type="text" name="description" /> <input type="file" name="file" /> <input type="submit" /> </form> However, I only get the file name, not the file content. When I add enctype="multipart/form-data" to the <form>, then request.getParameter() returns null. During research I stumbled upon Apache Common FileUpload. I tried this: FileItemFactory factory = new DiskFileItemFactory(); ServletFileUpload upload = new ServletFileUpload(factory); List items = upload.parseRequest(request); // This line is where it died. Unfortunately, the servlet threw an exception without a clear message and cause. Here is the stacktrace: SEVERE: Servlet.service() for servlet UploadServlet threw exception javax.servlet.ServletException: Servlet execution threw an exception at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:313) at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206) at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:233) at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:191) at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:127) at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102) at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109) at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:298) at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:852) at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:588) at org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:489) at java.lang.Thread.run(Thread.java:637)

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• What are some methods to prevent double posting in a form? (PHP)

  - by jpjp

  I want to prevent users from accidentally posting a comment twice. I use the PRG (post redirect get) method, so that I insert the data on another page then redirect the user back to the page which shows the comment. This allows users to refresh as many times as they want. However this doesn't work when the user goes back and clicks submit again or when they click submit 100 times really fast. I don't want 100 of the same comments. I looked at related questions on SO and found that a token is best. But I am having trouble using it. //makerandomtoken(20) returns a random 20 length char. <form method="post" ... > <input type="text" id="comments" name="comments" class="commentbox" /><br/> <input type="hidden" name="_token" value="<?php echo $token=makerandomtoken(20); ?>" /> <input type="submit" value="submit" name="submit" /> </form> if (isset($_POST['submit']) && !empty($comments)) { $comments= mysqli_real_escape_string($dbc,trim($_POST['comments'])); //how do I make the if-statment to check if the token has been already set once? if ( ____________){ //don't insert comment because already clicked submit } else{ //insert the comment into the database } } So I have the token as a hidden value, but how do I use that to prevent multiple clicking of submit. METHODS: someone suggested using sessions. I would set the random token to $_SESSION['_token'] and check if that session token is equal to the $_POST['_token'], but how do I do that? When I tried, it still doesn't check

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• Convert image color space and output separate channels in OpenCV

  - by Victor May

  I'm trying to reduce the runtime of a routine that converts an RGB image to a YCbCr image. My code looks like this: cv::Mat input(BGR->m_height, BGR->m_width, CV_8UC3, BGR->m_imageData); cv::Mat output(BGR->m_height, BGR->m_width, CV_8UC3); cv::cvtColor(input, output, CV_BGR2YCrCb); cv::Mat outputArr[3]; outputArr[0] = cv::Mat(BGR->m_height, BGR->m_width, CV_8UC1, Y->m_imageData); outputArr[1] = cv::Mat(BGR->m_height, BGR->m_width, CV_8UC1, Cr->m_imageData); outputArr[2] = cv::Mat(BGR->m_height, BGR->m_width, CV_8UC1, Cb->m_imageData); split(output,outputArr); But, this code is slow because there is a redundant split operation which copies the interleaved RGB image into the separate channel images. Is there a way to make the cvtColor function create an output that is already split into channel images? I tried to use constructors of the _OutputArray class that accepts a vector or array of matrices as an input, but it didn't work.

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• Array: Recursive problem cracked me up

  - by VaioIsBorn

  An array of integers A[i] (i 1) is defined in the following way: an element A[k] ( k 1) is the smallest number greater than A[k-1] such that the sum of its digits is equal to the sum of the digits of the number 4* A[k-1] . You need to write a program that calculates the N th number in this array based on the given first element A[1] . INPUT: In one line of standard input there are two numbers seperated with a single space: A[1] (1 <= A[1] <= 100) and N (1 <= N <= 10000). OUTPUT: The standard output should only contain a single integer A[N] , the Nth number of the defined sequence. Input: 7 4 Output: 79 Explanation: Elements of the array are as follows: 7, 19, 49, 79... and the 4th element is solution. I tried solving this by coding a separate function that for a given number A[k] calculates the sum of it's digits and finds the smallest number greater than A[k-1] as it says in the problem, but with no success. The first testing failed because of a memory limit, the second testing failed because of a time limit, and now i don't have any possible idea how to solve this. One friend suggested recursion, but i don't know how to set that. Anyone who can help me in any way please write, also suggest some ideas about using recursion/DP for solving this problem. Thanks.

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• Jquery unidentified data return when post

  - by Nhat Tuan

  <?php //load.php $myid = $_POST['id']; if($myid == 1){ echo '1'; }else if($myid == 0){ echo '0'; } ?> <html> <input id="id" /> <div id="result"></div> <input type="button" id="submit" /> <script type="text/javascript"> $('#submit').click( function(){ var send = 'id=' + $('#id').val(); $.post('load.php', send, function(data){ if(data == 1){ $('#result').html('Success'); }else if(data == 0){ $('#result').html('Failure'); }else{ $('#result').html('Unknow'); } }); }); </script> </html> I test this script in some free host and it work but in my real host jquery unidentified data return and it alway show 'Unknow'. When i change if(data == '1') it show 'Unknow' too EX: input id = 1 click submit & data return is 'Unknow' Why ?? I think this error from host, because i test it in some free host and it work, but now in my real host i got this error, how i can fix it ?

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• Bootstrap - Typehead on multiple inputs

  - by Clem

  I have two text intputs, both have to run an autocompletion. The site is using Bootstrap, and the « typeahead » component. I have this HTML : <input type="text" class="js_typeahead" data-role="artist" /> <input type="text" class="js_typeahead" data-role="location" /> I'm using the « data-role » attribute (that is sent to the Ajax controller as a $_POST index), in order to determine what kind of data has to be retrieved from the database. The Javascript goes this way : var myTypeahead = $('input.js_typeahead').typeahead({ source: function(query, process){ var data_role; data_role = myTypeahead.attr('data-role'); return $.post('/ajax/typeahead', { query:query,data_role:data_role },function(data){ return process(data.options); }); } }); With PHP, I check what $_POST['data-role'] contains, an run the MySQL query (in this case, a query either on a list of Artists, or a list of Locations). But the problem is the second "typeahead" returns the same values than the first one (list of Artists). I assume it's because the listener is attached to the object « myTypeahead », and this way the "data-role" attribute which is used, will always be the same. I think I could fix it by using something like : data_role = $(this).attr('data-role'); But of course this doesn't work, as it's a different scope. Maybe I'm doing it all wrong, but at least maybe you people could give me a hint. Sorry if this has already been discussed, I actually searched but without success. Thanks in advance, Clem (from France, sorry for my english)

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• How can I get a custom made set of checkboxes return values in the postback?

  - by AngryHacker

  I have the following in an aspx page: <td colspan="2"> <% DisplayParties(); %> </td> In the code behind for the aspx page, i have this (e.g. I build HTML for the checkboxes): public void DisplayParties() { var s = new StringBuilder(); s.Append("<input type=\"checkbox\" id=\"attorney\" value=\"12345\"/>"); s.Append("<input type=\"checkbox\" id=\"attorney\" value=\"67890\"/>"); s.Append("<input type=\"checkbox\" id=\"adjuster\" value=\"125\"/>"); Response.WriteLine(s.ToString()); } Not my proudest moment, but whatever. The problem is that when this page posts back via some event on the page, I never get these tags in the Request.Form collection. Is this simply how ASP.NET works (e.g. only server-side control post back) or am I missing something simple. My understanding was that a postback should bring back all the form variables.

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• Create / build / generate a web form that can be on my server and has modern looks and could be impl

  - by Luay

  I have a small web site and would like to add a 'contact us' form and a 'feedback' form. i would like the forms to satisfy the following: 1- be modern looking (with beautiful css effects) 2- the form fields are validated properly and 'inline'. What I mean is once a user skips a required field or enters an email address incorrectly some kind of tooltip or icon is displayed to ask him to cerrect the error (as opposed to a message box that appears after the user clicks 'submit') 3- once the submit button is clicked the form contents are emailed to me. 4- the whole thing can be setup by a noob like myself. 5- no ads on the form I have been searching for at least 5 days for a solution but I can't seem to find anything the would satisfy the above 5 conditions. I don't mind paying for a solution as long as it is hosted on my site and it is a one off payment and not a monthly payment. So far my search has lead me to the following: 1 wufoo. The good: the generated forms seem to look okay but not the best there is. The bad: the free service is limited to 100 submissions. ads on the form. it is not hosted on my server. Paid service requires monthly payments 2- emailmeform: almost same as above except the generated form looks old. They do have an offer where you pay only $4 to get the form and set it up on your own site but that doesn't solve the fact that the forms look old. 3- formAssembly: same as above with minor variations (the generated form looks better) 4- formchamp, formthis, kontaktr,... And other similar online services: the same problem. either the form generated looks outdated or require monthly payments or they put ads...they don't satisfy my conditions. 5- coffeecup form builder. a desktop software. The problem is the generated forms look too old and use flash. 6- simfatic. Another software. Much better than coffeecup. almost satisfies my conditions but the forms not as good as I like. 7- many, many php scripts or html templates that look so outdated or fail when tested (probably because they are too old). Seriously guys, how hard is this. At least 90%+ of website contain at least a 'contact us' form. Why aren't there better solutions? if there is I can't seem to find them. In terms of looks I want something similar to this: http://web-kreation.com/articles/lightform-free-ajaxphp-contact-form/ It is called lightform. And this is a perfect example of what I mean by 'inline' validation. the only problem: there is no script to handle sending the mail. Even if I find one, I don't know how to modify it for my needs. So could you please help me out. I really can't search anymore. I reached rock bottom with this issue. I need a complete solution. If nothing exists then at least a: 1- form template (html) that looks nice and can easily be modified 2- a validation script that does 'inline' validation like the example above (or similar to it) and can be easily implemented by a noob like me to work with the html form. 3- a php script that will handle sending the email and can be easily implemented (all three working in harmony). I hope there is a complete solution but am I asking for too much? Pretty please...help...

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• Problem with HiddenFor helper

  - by Dmitry Borovsky

  Hello. Model: public sealed class Model { public string Value { get; set; } } Controller: [HandleError] public class HomeController : Controller { [HttpGet] public ActionResult Index() { return View(new Model { Value = "+" } ); } [HttpPost] public ActionResult Index(Model model) { model.Value += "1"; return View(model); } } View: <%using (Html.BeginForm()){%> <%: Model.Value %> <%: Html.HiddenFor(model => model.Value) %> <input type="submit" value="ok"/> <%}%> Every time I submitted form result is <form action="/" method="post">+1 <input id="Value" name="Value" type="hidden" value="+"> <input type="submit" value="ok"> </form> It means that HiddenFor helper doesn't use real value of Model.Value but uses passed to controller one. Is it bug in MVC framework? Does anyone know workaround?

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• Displaying data from a linked table and displaying it as a list with HTML/PHP/MySQL

  - by user1672694

  I have three tables. students studentID | FirstName | LastName | Email | Form course CCode | Title courseenrolement courseenrolementid | studentID | ccode | complete | scode With the website, I have a page where I can view all the current enrolements and I wish to be able to view the list displaying the first name, surname and course title. I know I could do it with the following SQL (for the names): SELECT FirstName, LastName FROM student, courseenrolement WHERE courseenrolement.studentID = student.studentID But I am unsure how to get this to work using HTML/PHP. At present I only know how to display the studentID and CCode from the courseenrolement table using the following code: <ul> <?php foreach ($courseenrolements as $ce): ?> <li> <form action="" method="post"> <div> <?php htmlout($ce['studentID']); ?> <?php htmlout($ce['CCode']); ?> <input type="hidden" name="courseenrolementid" value="<?php echo $ce['courseenrolementid']; ?>"> <input type="submit" name="action" value="Edit"> <input type="submit" name="action" value="Delete"> </div> </form> </li> <?php endforeach; ?> </ul> which displays this: But I would like the names and course title. I managed to get it to show the names etc in the dropdown on the 'Add new' form, so I would assume it will be similar, but just unsure on how exactly. Thanks in advance

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• Using C# to HttpPost data to a web page

  - by druffmuff

  I want to log in into a website using C# code. Here's the html code of the example form: <form action="http://www.site.com/login.php" method="post" name="login" id="login"> <table border="0" cellpadding="2" cellspacing="0"> <tbody> <tr><td><b>User:</b></td><td colspan=\"2\"><b>Password:</b></td></tr> <tr> <td><input class="inputbg" name="user" type="text"></td> <td><input class="inputbg" name="password" type="password"></td> <td><input type="submit" name="user_control" value="Submit" class="buttonbg"></td> </tr> </tbody></table> </form> This is what I have tried so far with unsuccessful results: HttpWebRequest request = (HttpWebRequest)WebRequest.Create("http://www.site.com/login.php"); request.Method = "POST"; using (StreamWriter writer = new StreamWriter(request.GetRequestStream(), Encoding.ASCII)) { writer.Write("user=user&password=pass&user_control=Eingabe"); } HttpWebResponse response = (HttpWebResponse)request.GetResponse(); using (StreamReader reader = new StreamReader(response.GetResponseStream())) { stream = new StreamWriter("login.html"); stream.Write(reader.ReadToEnd()); stream.Close(); } Any Ideas, why this is failing?

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• Strange issues with view switcher after object animator animations

  - by Barry Irvine

  I have two LinearLayout views that contain a number of edit texts and checkboxes for entering user information (name, email address etc). When a validation fails on one of these fields a gone textview is displayed showing the validation error. I have enclosed the two layouts within a ViewSwitcher and I animate between the two views using the ObjectAnimator class. (Since the code needs to support older versions of Android I am actually using the nineoldandroids backwards compatibility library for this). The bulk of the work is performed in my switchToChild method. If I flip the views more than twice then I start to run into strange errors. Firstly although the correct child view of the view animator is displayed it seems that the other view has focus and I can click on the views beneath the current one. I resolved this issue by adding a viewSwitcher.bringChildToFront at the end of the first animation. When I do this however and perform a validation on the 2nd view the "gone" view that I have now set to visible is not displayed (as if the linearlayout is never being re-measured). Here is a subset of the XML file: <ScrollView android:layout_width="fill_parent" android:layout_height="wrap_content" android:layout_below="@+id/TitleBar" android:scrollbarAlwaysDrawVerticalTrack="true" android:scrollbarStyle="outsideOverlay" android:scrollbars="vertical" > <ViewSwitcher android:id="@+id/switcher" android:layout_width="fill_parent" android:layout_height="wrap_content" > <LinearLayout android:id="@+id/page_1" android:layout_width="fill_parent" android:layout_height="wrap_content" android:orientation="vertical" > <!-- Lots of subviews here --> <LinearLayout android:id="@+id/page_2" android:layout_width="fill_parent" android:layout_height="wrap_content" android:orientation="vertical" > And this is the main method for flipping between the views: private void switchToChild(final int child) { final ViewSwitcher viewSwitcher = (ViewSwitcher) findViewById(R.id.switcher); if (viewSwitcher.getDisplayedChild() != child) { final Interpolator accelerator = new AccelerateInterpolator(); final Interpolator decelerator = new DecelerateInterpolator(); final View visibleView; final View invisibleView; switch (child) { case 0: visibleView = findViewById(R.id.page_2); invisibleView = findViewById(R.id.page_1); findViewById(R.id.next).setVisibility(View.VISIBLE); findViewById(R.id.back).setVisibility(View.GONE); break; case 1: default: visibleView = findViewById(R.id.page_1); invisibleView = findViewById(R.id.page_2); findViewById(R.id.back).setVisibility(View.VISIBLE); findViewById(R.id.next).setVisibility(View.GONE); break; } final ObjectAnimator visToInvis = ObjectAnimator.ofFloat(visibleView, "rotationY", 0f, 90f).setDuration(250); visToInvis.setInterpolator(accelerator); final ObjectAnimator invisToVis = ObjectAnimator.ofFloat(invisibleView, "rotationY", -90f, 0f).setDuration(250); invisToVis.setInterpolator(decelerator); visToInvis.addListener(new AnimatorListenerAdapter() { @Override public void onAnimationEnd(Animator anim) { viewSwitcher.showNext(); invisToVis.start(); viewSwitcher.bringChildToFront(invisibleView); // If I don't do this the old view can have focus } }); visToInvis.start(); } } Does anyone have any ideas? This is really confusing me!

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